这几道题，麻烦您给解一下

1,4是微分方程吗?我都忘了,去问别人吧,抱歉.

2.(cos2x)^(1/x²)=e^[(1/x²)lncos2x],然后用等价无穷小来替代,

lncos2x=ln[1+cos2x-1]∽cos2x-1,[∵lim(t→0)[ln(1+t)]/t=lim(t→0)ln(1+t)^(1/t)=lne=1,∴ln(1+t)∽t];

cos2x-1=-2sin²x∽-2x²[∵-2sin²x/(-2x²)=(sinx/x)²=1,x→0],∴lncos2x∽-2x²,

∴lim(x→0)(cos2x)^(1/x²)=lim(x→0)e^[(1/x²)lncos2x]=lim(x→0)e^[(1/x²)*(-2x²)]=1/e²;

3.lim(x→0)(sinx/x)^(1/x)=e^lim(x→0)[ln(sinx/x)]/x=(2次使用洛必达法则)=e^0=1.